Vertical limit functions

We will now introduce the concept of a vertical limit function, which will be of great impor-tance in the study of composition operators onH ². In order to do so, we need the notion of a multiplicative character. We say that χ:N→T is a multiplicative character if it satisfies χ(mn) =χ(m)χ(n) for all positive integersmandn. We denote the set of all such characters by M. It is convenient to identify the set M with T^∞, the infinite dimensional Cartesian product of T. This identification is done as follows. Take a point z = (z₁, z₂, ...) ∈ T^∞ and let χ(p_j) = z_j, where p_j denotes the j-th prime. The character χ is now defined for every prime number and we extend it to all of N by letting χ(n) =χ(p^r₁¹)· · ·χ(p^r_m^m), where n = p^r₁¹ · · · p^r_m^m. Note that this definition of χ forces it to be multiplicative. Moreover, T^∞ is a compact group under point-wise multiplication. It therefore exists a unique Haar measure on T^∞, which happens to coincide with the infinite product measure generated by the normalized Lebesgue measure on T.

Consider now the function f(s) = P∞

n=1a_nn^−s ∈ D. For any characterχ∈ M we define the function fχ by

f_χ(s) =

∞

X

n=1

a_nχ(n)n^−s. (2.2)

We call this a vertical limit function. To understand the reasoning behind this name we look at the character χ(n) = n^−it. This obviously defines a multiplicative character. Also, for f(s) =P∞

n=1a_nn^−s we havef_χ(s) =f(s+iτ) =:f_τ(s). That is, f_χ is a vertical translation of f. The interesting thing is, as we shall see shortly, that every vertical limit function fχ

corresponds to the limit of a sequence of vertical translations of f. We start out by proving a famous theorem by Kronecker, and we will give an analytic proof due to Bohr [3].

Definition 2.6. Let a1, ..., an be a set of real numbers. We say that the numbers are Q -linearly independent if

n

X

j=1

cjaj = 0, with c1, ..., cn ∈Z, only when c_j = 0 for 1≤j ≤n.

One typical example of such a set is the following. If p₁, ..., p_j is a set ofj unique primes, then the real numbers logp₁, ...,logp_j are Q-linearly independent. This follows from the relation

j

X

i=1

c_ilogp_i = log Y

1≤i≤j

p^c_iⁱ, and the fundamental theorem of arithmetic.

Theorem 2.7 (Kronecker’s Theorem). Suppose ξ1, ..., ξk are Q-linearly independent real numbers. Let α₁, ..., α_k be arbitrary real numbers and let > 0. Then there exists integers n₁, ..., n_k and a real number t satisfying

|tξj −αj −nj|< , j= 1,2, ..., k.

Proof. We want to carry out an analytic proof of this result and we will rely on the following fact. The exponential function e^2πix is equal to 1 if and only if xis an integer. Consider the function to an integer for all m. In this language, Kronecker’s Theorem states that we can get |f(t)|

arbitrarily close to k+ 1 by choosing t sufficiently large.

Let F denote the function

F(x1, .., xk) = 1 +x1+· · ·+xk.

If we now raise the functions f and F to their p-th power, for some integer p, and use the multinomial theorem, they take the form

(f(t))^p =X

bνe^iβνt, (F(x₁, ..., x_k))^p =X

b_λ1,...,λkx^λ₁¹ · · ·x^λ_k^k.

Note that the β_ν must all be different due to the linear independence of the ξ_i. The p-th power of the functions f and F must therefore have the same number of terms, and the absolute value of the coefficients b_ν and b_λ1,...,λk coincide. This yields

X|bν|=X

bλ1,...,λ_k = (F(1, ...,1))^p = (k+ 1)^p. For any fixed ν =ν⁰ we have the following identity,

Tlim→∞ then every coefficient b_ν of (f(t))^p satisfies

|b_ν|=

Askincreases by 1, the number of terms in the polynomial development of (1 +x₁+...+x_k)^p is multiplied with at most p+ 1. This means that the total number of terms must be less than (p+ 1)^k. We therefore have

X|b_ν|<(p+ 1)^kC^p, which implies that

(p+ 1)^kC^p

P|bν| = (p+ 1)^kC^p (k+ 1)^p >1 for all p. However, since C < k+ 1 we have that

(p+ 1)^kC^p (k+ 1)^p →0,

as p → ∞. Therefore, our assumption that |f(t)| is less than C for every t ∈ R leads to a contradiction. Consequently, the proof is complete.

The result below was originally presented in [10], and we will follow a proof from [13].

Theorem 2.8. Suppose that f(s) = P

n=1a_nn^−s, with σ_u(f) = θ. Then the vertical limit functions f_χ are precisely the limits of some sequence {f_τN} of vertical translations of f, in the half-plane Cθ. The sequence {f_τN} converges uniformly on Cθ+δ, for every δ > 0.

Moreover, we have σ_u(f_χ) = θ.

Proof. Assume that we have a sequence of vertical translations {f_τN}N≥1 converging to an analytic function on Cθ. It is clear that the function defined as ξ_τN(n) := n^−iτN must converge to some limit functionξ(n), as N → ∞. We readily see that |ξ(n)|= 1 and that ξ is completely multiplicative. Hence, there exists a character χ ∈ M such that χ(n) =ξ(n) for alln. Letϑ > θ. We know thatf converges uniformly on Cϑ. We can then take the limit as N approaches infinity of the translations f_τN, and pass the limit inside the summation.

This will, in conjunction with the argument above, ensure that f_τN converges to f_χ on Cϑ. Since ϑ was arbitrary, we actually have convergence on all of Cθ and alsoσ_u(f_χ) = θ.

On the other hand, we want to show that for any characterχ∈ Mthere exists a sequence of real numbers {τ_N}N≥1 such that {f_τN}N≥1 converges to f_χ. For this purpose, we recall Kronecker’s Theorem (Theorem 2.7). In the language of vertical limit functions the theorem says that there exists τ ∈R such that

|χ(n)−n^−iτ|< ε, n= 1, ..., N,

for anyε >0 and positive integer N. Then for allN we should be able to findτ_N such that

|χ(n)−n^−iτN|<1/N n= 1, ..., N.

As N approach infinity we see thatn^−iτN →χ(n), which is what we wanted to prove.

Since all characters χ has modulus 1 it is also clear that σ_a(f_χ) = σ_a(f). We will now establish a product formula for vertical limit functions. The lemma is a slight improvement of a result from [13].

Lemma 2.9. Assume we have a characterχ∈ M and two convergent Dirichlet seriesf and g. Then the product formula (f g)_χ =f_χg_χ holds in Cθ whenever σ_u(f)≤θ and σ_u(g)≤θ.

Proof. In a remote half-plane where both f and g converges absolutely we can write the product f(s)g(s) = (P∞

n=1a_nn^−s) (P∞

m=1b_mm^−s) as f(s)g(s) =

∞

X

n=1

c_nn^−s, where

c_n= X

ij=n

a_ib_j. Then

(f g)_χ=

∞

X

n=1

c_nχ(n)n^−s. On the other hand we have

f_χg_χ=

∞

X

n=1

d_nn^−s, where

d_n= X

ij=n

a_ib_jχ(i)χ(j).

But χ is a completely multiplicative character, so dn =χ(n)X

ij=n

aibj =χ(n)cn.

We see that (f g)_χ =f_χg_χin a half-plane wheref andg converges absolutely. Sinceσ_u(f)≤θ andσ_u(g)≤θ, the two functions are bounded onCθ. The product of two bounded functions are again bounded, so by Bohr’s theorem it follows that σ_u(f g) ≤ θ. The product formula therefore holds in all ofCθ.

The next result tells us that the image of a function is invariant under vertical translations.

We follow a proof from [5].

Lemma 2.10. Let ψ : Cθ → Cν be analytic, with ψ(s) = P∞

n=1c_nn^−s. Then ψ(Cθ) = ψχ(C^θ).

Proof. We observe thatψ(+∞) = ψ_χ(∞) =c₁, so the result is clear for constantψ. Assume now that ψ is non-constant. For a point w∈ Cθ we find a closed disc K which contains w in its interior and satisfies

M = inf

s∈∂K|ψ_χ(s)−ψ_χ(w)|>0.

By Theorem 2.8 there exists a sequence of vertical translation of ψ converging to ψχ on K.

Let τ_k be a sequence such that ψ(s+iτ_k) → ψ_χ(s). By choosing k large enough we can ensure that

|ψ(s+iτk)−ψχ|< M.

We add and subtract ψ_χ(w) on the left-hand side, so that

|ψ(s+iτ_k)−ψ_χ(w)−(ψ_χ(s)−ψ_χ(w))|< M, valid for s ∈K. Denote by f and g the functions

f(s) = ψ_χ(s)−ψ_χ(w)

g(s) = ψ(s+iτ_k)−ψ_χ(w)−(ψ_χ(s)−ψ_χ(w)).

The function f is clearly zero at s=w. Rouch´e’s theorem tells us that f and f+g has the same number so zeros in K. We have

(f +g)(s) =ψ(s+iτ_k)−ψ_χ(w),

so there must be a value of s inK makingψ(s+iτ_k) =ψ_χ(w), and the proof is done.

Chapter 3

In document Composition operators on the Hardy space of Dirichlet series (sider 36-41)